363777
What is the power output of \(_{92}{U^{235}}\) reactor if it takes 30 days to use up \(2g\) of fuel and if each fission gives \(185\,MeV\) of usable energy? Avogadro’ number \( = 6.02 \times {10^{26}}mo{\ell ^{ - 1}}\)
1 \(45\,MW\)
2 \(58.46\,MW\)
3 \(72\,MW\)
4 \(92\,MW\)
Explanation:
Number of atoms in \(2kg\) fuel \(\frac{2}{{235}} \times 6.02 \times {10^{26}} = 5.12 \times {10^2}\) Number of atoms fissioned in one second \( = \frac{{5.12 \times {{10}^{24}}}}{{30 \times 24 \times 60 \times 60}}\) \( = 1.975 \times {10^{18}}{s^{ - 1}}\) Each fission gives \(185\,MeV\). Hence, energy obtained in one second, \(P = 185 \times 1.975 \times {10^{18}}MeV{s^{ - 1}}\) \( = 185 \times 1.975 \times {10^{18}} \times 1.6 \times {10^{ - 19}}j{s^{ - 1}}\) \( = 58.46\) Megawatt
PHXII13:NUCLEI
363778
On bombarding \({U^{235}}\) by slow neutron, 200 \(MeV\) energy is released. If the power output of atomic rector is 1.6 \(MW\), then the rate of fission will be
1 \(8 \times {10^{16}}/s\)
2 \(20 \times {10^{16}}/s\)
3 \(5 \times {10^{22}}/s\)
4 \(5 \times {10^{16}}/s\)
Explanation:
Energy released per fission of uranium nucleus \( = 200 \times {10^6} \times 1.6 \times {10^{ - 19}}J\) Power output \( = 1.6 \times {10^6}\) \(\therefore \) Number of fission/s \( = \frac{{{\rm{Power}}\,{\rm{output}}}}{{{\rm{Energy}}\,{\rm{release}}\,{\rm{per}}\,{\rm{fission}}}}\) \( = \frac{{1.6 \times {{10}^6}}}{{200 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}} = 5 \times {10^{16}}/s\) This is the rate of fission
KCET - 2008
PHXII13:NUCLEI
363779
In a nuclear reactor using \(U^{235}\), the power output is \(9.6\,MW\). If energy released per fission of \(U^{235}\) is \(200\,MeV\), then the number of fissions per second is (\(1\,eV = 1.6 \times {10^{ - 19}}J\))
363780
In a nuclear reactor heavy nuclei is not used as moderators because
1 They will break up
2 Elastic collision of neutrons with heavy nuclei will not slow them down
3 The net weight of the reactor would be unbearably high
4 Substance with heavy nuclei do not occur in liquid or gaseous state at room temperature
Explanation:
Since the function of a moderator is to reduce the speed of neutrons, materials with lighter nuclei are chosen for this purpose so as to maximise the reduction in kinetic energy of neutrons. For a near-elastic collision with heavy nucleus, the speed of a neutron may not reduce much.
363777
What is the power output of \(_{92}{U^{235}}\) reactor if it takes 30 days to use up \(2g\) of fuel and if each fission gives \(185\,MeV\) of usable energy? Avogadro’ number \( = 6.02 \times {10^{26}}mo{\ell ^{ - 1}}\)
1 \(45\,MW\)
2 \(58.46\,MW\)
3 \(72\,MW\)
4 \(92\,MW\)
Explanation:
Number of atoms in \(2kg\) fuel \(\frac{2}{{235}} \times 6.02 \times {10^{26}} = 5.12 \times {10^2}\) Number of atoms fissioned in one second \( = \frac{{5.12 \times {{10}^{24}}}}{{30 \times 24 \times 60 \times 60}}\) \( = 1.975 \times {10^{18}}{s^{ - 1}}\) Each fission gives \(185\,MeV\). Hence, energy obtained in one second, \(P = 185 \times 1.975 \times {10^{18}}MeV{s^{ - 1}}\) \( = 185 \times 1.975 \times {10^{18}} \times 1.6 \times {10^{ - 19}}j{s^{ - 1}}\) \( = 58.46\) Megawatt
PHXII13:NUCLEI
363778
On bombarding \({U^{235}}\) by slow neutron, 200 \(MeV\) energy is released. If the power output of atomic rector is 1.6 \(MW\), then the rate of fission will be
1 \(8 \times {10^{16}}/s\)
2 \(20 \times {10^{16}}/s\)
3 \(5 \times {10^{22}}/s\)
4 \(5 \times {10^{16}}/s\)
Explanation:
Energy released per fission of uranium nucleus \( = 200 \times {10^6} \times 1.6 \times {10^{ - 19}}J\) Power output \( = 1.6 \times {10^6}\) \(\therefore \) Number of fission/s \( = \frac{{{\rm{Power}}\,{\rm{output}}}}{{{\rm{Energy}}\,{\rm{release}}\,{\rm{per}}\,{\rm{fission}}}}\) \( = \frac{{1.6 \times {{10}^6}}}{{200 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}} = 5 \times {10^{16}}/s\) This is the rate of fission
KCET - 2008
PHXII13:NUCLEI
363779
In a nuclear reactor using \(U^{235}\), the power output is \(9.6\,MW\). If energy released per fission of \(U^{235}\) is \(200\,MeV\), then the number of fissions per second is (\(1\,eV = 1.6 \times {10^{ - 19}}J\))
363780
In a nuclear reactor heavy nuclei is not used as moderators because
1 They will break up
2 Elastic collision of neutrons with heavy nuclei will not slow them down
3 The net weight of the reactor would be unbearably high
4 Substance with heavy nuclei do not occur in liquid or gaseous state at room temperature
Explanation:
Since the function of a moderator is to reduce the speed of neutrons, materials with lighter nuclei are chosen for this purpose so as to maximise the reduction in kinetic energy of neutrons. For a near-elastic collision with heavy nucleus, the speed of a neutron may not reduce much.
363777
What is the power output of \(_{92}{U^{235}}\) reactor if it takes 30 days to use up \(2g\) of fuel and if each fission gives \(185\,MeV\) of usable energy? Avogadro’ number \( = 6.02 \times {10^{26}}mo{\ell ^{ - 1}}\)
1 \(45\,MW\)
2 \(58.46\,MW\)
3 \(72\,MW\)
4 \(92\,MW\)
Explanation:
Number of atoms in \(2kg\) fuel \(\frac{2}{{235}} \times 6.02 \times {10^{26}} = 5.12 \times {10^2}\) Number of atoms fissioned in one second \( = \frac{{5.12 \times {{10}^{24}}}}{{30 \times 24 \times 60 \times 60}}\) \( = 1.975 \times {10^{18}}{s^{ - 1}}\) Each fission gives \(185\,MeV\). Hence, energy obtained in one second, \(P = 185 \times 1.975 \times {10^{18}}MeV{s^{ - 1}}\) \( = 185 \times 1.975 \times {10^{18}} \times 1.6 \times {10^{ - 19}}j{s^{ - 1}}\) \( = 58.46\) Megawatt
PHXII13:NUCLEI
363778
On bombarding \({U^{235}}\) by slow neutron, 200 \(MeV\) energy is released. If the power output of atomic rector is 1.6 \(MW\), then the rate of fission will be
1 \(8 \times {10^{16}}/s\)
2 \(20 \times {10^{16}}/s\)
3 \(5 \times {10^{22}}/s\)
4 \(5 \times {10^{16}}/s\)
Explanation:
Energy released per fission of uranium nucleus \( = 200 \times {10^6} \times 1.6 \times {10^{ - 19}}J\) Power output \( = 1.6 \times {10^6}\) \(\therefore \) Number of fission/s \( = \frac{{{\rm{Power}}\,{\rm{output}}}}{{{\rm{Energy}}\,{\rm{release}}\,{\rm{per}}\,{\rm{fission}}}}\) \( = \frac{{1.6 \times {{10}^6}}}{{200 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}} = 5 \times {10^{16}}/s\) This is the rate of fission
KCET - 2008
PHXII13:NUCLEI
363779
In a nuclear reactor using \(U^{235}\), the power output is \(9.6\,MW\). If energy released per fission of \(U^{235}\) is \(200\,MeV\), then the number of fissions per second is (\(1\,eV = 1.6 \times {10^{ - 19}}J\))
363780
In a nuclear reactor heavy nuclei is not used as moderators because
1 They will break up
2 Elastic collision of neutrons with heavy nuclei will not slow them down
3 The net weight of the reactor would be unbearably high
4 Substance with heavy nuclei do not occur in liquid or gaseous state at room temperature
Explanation:
Since the function of a moderator is to reduce the speed of neutrons, materials with lighter nuclei are chosen for this purpose so as to maximise the reduction in kinetic energy of neutrons. For a near-elastic collision with heavy nucleus, the speed of a neutron may not reduce much.
363777
What is the power output of \(_{92}{U^{235}}\) reactor if it takes 30 days to use up \(2g\) of fuel and if each fission gives \(185\,MeV\) of usable energy? Avogadro’ number \( = 6.02 \times {10^{26}}mo{\ell ^{ - 1}}\)
1 \(45\,MW\)
2 \(58.46\,MW\)
3 \(72\,MW\)
4 \(92\,MW\)
Explanation:
Number of atoms in \(2kg\) fuel \(\frac{2}{{235}} \times 6.02 \times {10^{26}} = 5.12 \times {10^2}\) Number of atoms fissioned in one second \( = \frac{{5.12 \times {{10}^{24}}}}{{30 \times 24 \times 60 \times 60}}\) \( = 1.975 \times {10^{18}}{s^{ - 1}}\) Each fission gives \(185\,MeV\). Hence, energy obtained in one second, \(P = 185 \times 1.975 \times {10^{18}}MeV{s^{ - 1}}\) \( = 185 \times 1.975 \times {10^{18}} \times 1.6 \times {10^{ - 19}}j{s^{ - 1}}\) \( = 58.46\) Megawatt
PHXII13:NUCLEI
363778
On bombarding \({U^{235}}\) by slow neutron, 200 \(MeV\) energy is released. If the power output of atomic rector is 1.6 \(MW\), then the rate of fission will be
1 \(8 \times {10^{16}}/s\)
2 \(20 \times {10^{16}}/s\)
3 \(5 \times {10^{22}}/s\)
4 \(5 \times {10^{16}}/s\)
Explanation:
Energy released per fission of uranium nucleus \( = 200 \times {10^6} \times 1.6 \times {10^{ - 19}}J\) Power output \( = 1.6 \times {10^6}\) \(\therefore \) Number of fission/s \( = \frac{{{\rm{Power}}\,{\rm{output}}}}{{{\rm{Energy}}\,{\rm{release}}\,{\rm{per}}\,{\rm{fission}}}}\) \( = \frac{{1.6 \times {{10}^6}}}{{200 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}} = 5 \times {10^{16}}/s\) This is the rate of fission
KCET - 2008
PHXII13:NUCLEI
363779
In a nuclear reactor using \(U^{235}\), the power output is \(9.6\,MW\). If energy released per fission of \(U^{235}\) is \(200\,MeV\), then the number of fissions per second is (\(1\,eV = 1.6 \times {10^{ - 19}}J\))
363780
In a nuclear reactor heavy nuclei is not used as moderators because
1 They will break up
2 Elastic collision of neutrons with heavy nuclei will not slow them down
3 The net weight of the reactor would be unbearably high
4 Substance with heavy nuclei do not occur in liquid or gaseous state at room temperature
Explanation:
Since the function of a moderator is to reduce the speed of neutrons, materials with lighter nuclei are chosen for this purpose so as to maximise the reduction in kinetic energy of neutrons. For a near-elastic collision with heavy nucleus, the speed of a neutron may not reduce much.
